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## More problems from Liboff chapter 4. Hermitian operators.

Posted by peeterjoot on June 27, 2010

# Motivation.

Some more problems from [1].

# Problem 4.11

Some problems on Hermitian adjoints. The starting point is the definition of the adjoint $A^\dagger$ of $A$ in terms of the inner product

\begin{aligned}\left\langle{{\hat{A}^\dagger \phi}} \vert {{\psi}}\right\rangle = \left\langle{{\phi}} \vert {{\hat{A} \psi}}\right\rangle\end{aligned}

## 4.11 a

\begin{aligned}\left\langle{{ \phi }} \vert {{ (a \hat{A} + b \hat{B}) \psi }}\right\rangle &=a \left\langle{{ \phi }} \vert {{ \hat{A} \psi }}\right\rangle + b \left\langle{{ \phi }} \vert {{ \hat{B} \psi }}\right\rangle \\ &=a \left\langle{{ \hat{A}^\dagger \phi }} \vert {{ \psi }}\right\rangle + b \left\langle{{ \hat{B}^\dagger \phi }} \vert {{ \psi }}\right\rangle \\ &=\left\langle{{ a^{*} \hat{A}^\dagger \phi }} \vert {{ \psi }}\right\rangle + \left\langle{{ b^{*} \hat{B}^\dagger \phi }} \vert {{ \psi }}\right\rangle \\ &=\left\langle{{ (a^{*} \hat{A}^\dagger + b^{*} \hat{B}^\dagger ) \phi }} \vert {{ \psi }}\right\rangle \\ &\implies \\ (a \hat{A} + b \hat{B})^\dagger = (a^{*} \hat{A}^\dagger + b^{*} \hat{B}^\dagger)\end{aligned}

## 4.11 b

\begin{aligned}\left\langle{{ \phi }} \vert {{ \hat{A} \hat{B} \psi }}\right\rangle &=\left\langle{{ \hat{A}^\dagger \phi }} \vert {{ \hat{B} \psi }}\right\rangle \\ &=\left\langle{{ \hat{B}^\dagger \hat{A}^\dagger \phi }} \vert {{ \psi }}\right\rangle \\ &\implies \\ (\hat{A} \hat{B} )^\dagger &=\hat{B}^\dagger \hat{A}^\dagger \end{aligned}

## 4.11 d

Hermitian adjoint of $D^2$, where $D = {\partial {}}/{\partial {x}}$. Here we need the integral form of the inner product

\begin{aligned}\left\langle{{\phi}} \vert {{D^2 \psi}}\right\rangle &=\int \phi^{*} \frac{\partial {}}{\partial {x}}\frac{\partial {\psi}}{\partial {x}} \\ &=-\int \frac{\partial {\phi^{*}}}{\partial {x}} \frac{\partial {\psi}}{\partial {x}} \\ &=\int \psi \frac{\partial {}}{\partial {x}}\frac{\partial {\phi^{*}}}{\partial {x}} \\ &\implies \\ (D^2)^\dagger &= D^2\end{aligned}

Since the text shows that the square of a Hermitian operator is Hermitian, one perhaps wonders if $D$ is (but we expect not since $\hat{p} = -i \hbar D$ is Hermitian).

Suppose $\hat{A} = aD$, we have

\begin{aligned}\hat{A}^\dagger = -a^{*} D,\end{aligned}

so for this to be Hermitian ($\hat{A} = \hat{A}^\dagger$) we must have $- a^{*} = a$. If $a = r e^{i\theta}$, we have

\begin{aligned}-1 = e^{2 i\theta}\end{aligned}

So $\theta = \pi (1/2 + n)$, and $a = \pm i r$. This fixes the scalar multiples of $D$ that are required to form a Hermitian operator

\begin{aligned}\hat{A} &= \pm i r D\end{aligned}

where $r$ is any real positive constant.

## 4.11 e

\begin{aligned}(\hat{A} \hat{B} - \hat{B} \hat{A})^\dagger &= - (\hat{A}^\dagger \hat{B}^\dagger - \hat{B}^\dagger \hat{A}^\dagger)\end{aligned}

## 4.11 f

\begin{aligned}(\hat{A} \hat{B} + \hat{B} \hat{A})^\dagger &= \hat{A}^\dagger \hat{B}^\dagger + \hat{B}^\dagger \hat{A}^\dagger\end{aligned}

## 4.11 g

\begin{aligned}i (\hat{A} \hat{B} - \hat{B} \hat{A})^\dagger &= i ( \hat{A}^\dagger \hat{B}^\dagger - \hat{B}^\dagger \hat{A}^\dagger)\end{aligned}

## 4.11 h

This one was to calculate $(\hat{A}^\dagger)^\dagger$. Intuitively I’d expect that $(\hat{A}^\dagger)^\dagger = \hat{A}$. How could one show this?

Trying to show this with Dirac notation, I got all mixed up initially.

Using the more straightforward and old fashioned integral notation (as in [2]), this is more straightforward. We have the Hermitian conjugate defined by

\begin{aligned}\int \psi_2^{*} (\hat{A} \psi_1) = \int (\hat{A}^\dagger \psi_2^{*}) \psi_1,\end{aligned}

Or, more symmetrically, using braces to indicate operator direction

\begin{aligned}\int \psi_2^{*} (\hat{A} \psi_1) = \int (\psi_2^{*} \hat{A}^\dagger) \psi_1.\end{aligned}

Introduce a couple of variable substuitions for clarity

\begin{aligned}\phi_1 &= \psi_1^{*} \\ \phi_2 &= \psi_2^{*} \\ \hat{B} &= \hat{A}^\dagger.\end{aligned}

We then have

\begin{aligned}\int \psi_2^{*} (\hat{A} \psi_1)&=\int (\psi_2^{*} \hat{A}^\dagger) \psi_1 \\ &=\int (\phi_2 \hat{B}) \phi_1^{*} \\ &=\int \phi_1^{*} (\hat{B} \phi_2) \\ &=\int (\phi_1^{*} \hat{B}^\dagger) \phi_2 \\ &=\int \phi_2 (\hat{B}^\dagger \phi_1^{*}) \\ &=\int \psi_2^{*} (\hat{A}^{\dagger \dagger} \psi_1) \\ \end{aligned}

Since this is true for all $\psi_k$, we have $\hat{A} = \hat{A}^{\dagger \dagger}$ as expected.

Having figured out the problem in the simpleton way, it’s now simple to go back and translate this into the Dirac inner product notation without getting muddled. We have

\begin{aligned}\left\langle{{ \psi_2 }} \vert {{ \hat{A} \psi_1 }}\right\rangle &=\left\langle{{ \hat{A}^\dagger \psi_2 }} \vert {{ \psi_1 }}\right\rangle \\ &=\left\langle{{ \hat{B} \phi_2^{*} }} \vert {{ \phi_1^{*} }}\right\rangle \\ &={\left\langle{{ \phi_1 }} \vert {{ \hat{B}^{*} \phi_2}}\right\rangle}^{*} \\ &={\left\langle{{ (\hat{B}^{*})^\dagger \phi_1 }} \vert {{ \phi_2}}\right\rangle}^{*} \\ &=\left\langle{{\phi_2^{*} }} \vert {{ \hat{B}^\dagger \phi_1^{*} }}\right\rangle \\ &=\left\langle{{\psi_2 }} \vert {{ \hat{A}^{\dagger \dagger} \psi_1 }}\right\rangle \\ \end{aligned}

## 4.11 i

\begin{aligned}(\hat{A} \hat{A}^\dagger)^\dagger &= (\hat{A}^\dagger)^\dagger \hat{A}^\dagger \end{aligned}

since $(\hat{A}^\dagger) ^\dagger = \hat{A}$

\begin{aligned}(\hat{A} \hat{A}^\dagger)^\dagger &= \hat{A} \hat{A}^\dagger.\end{aligned}

# Problem 4.12 d

If $\hat{A}$ is not Hermitian, is the product $\hat{A}^\dagger \hat{A}$ Hermitian? To start we need to verify that $\left\langle{{\psi}} \vert {{\hat{A}^\dagger \phi}}\right\rangle = \left\langle{{\hat{A} \psi}} \vert {{\phi}}\right\rangle$.

\begin{aligned}\left\langle{{ \psi }} \vert {{ \hat{A}^\dagger \phi }}\right\rangle &={\left\langle{{ (\hat{A}^\dagger)^{*} \phi^{*} }} \vert {{ \psi^{*} }}\right\rangle}^{*} \\ &={\left\langle{{ \phi^{*} }} \vert {{ \hat{A}^{*} \psi^{*} }}\right\rangle}^{*} \\ &=\left\langle{{ \psi }} \vert {{ \hat{A} \psi }}\right\rangle.\end{aligned}

With that verified we have

\begin{aligned}\left\langle{{ \psi }} \vert {{ \hat{A}^\dagger \hat{A} \phi }}\right\rangle &=\left\langle{{ \hat{A} \psi }} \vert {{ \hat{A} \phi }}\right\rangle \\ &=\left\langle{{ \hat{A}^\dagger \hat{A} \psi }} \vert {{ \phi }}\right\rangle,\end{aligned}

so, the answer is yes. Provided the adjoint exists, that product will be Hermitian.

# Problem 4.14

Show that $\left\langle{{\hat{A}}}\right\rangle = \left\langle{{\hat{A}}}\right\rangle^{*}$ (that it is real), if $\hat{A}$ is Hermitian. This follows by expansion of that conjuagate

\begin{aligned}\left\langle{{\hat{A}}}\right\rangle^{*} &= \left(\int \psi^{*} \hat{A} \psi \right)^{*} \\ &= \int \psi \hat{A}^{*} \psi^{*} \\ &= \int (\hat{A} \psi)^{*} \psi \\ &= \left\langle{{ \hat{A} \psi }} \vert {{ \psi }}\right\rangle \\ &= \left\langle{{ \psi }} \vert {{ \hat{A}^\dagger \psi }}\right\rangle \\ &= \left\langle{{ \psi }} \vert {{ \hat{A} \psi }}\right\rangle \\ &= \left\langle{{\hat{A}}}\right\rangle\end{aligned}

# References

[1] R. Liboff. Introductory quantum mechanics. 2003.

[2] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.